Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.5 Solutions: Introduction to Linear Polynomials

Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.5 Solutions cover linear relationships from Introduction to Linear Polynomials. This exercise helps students find the values of a and b in equations of the form y = ax + b.

In Ganita Manjari Class 9 Chapter 2 Exercise 2.5, students work with monthly bills, module access, gym charges, badminton court usage and temperature conversion. These questions show how a linear relationship Class 9 can be formed when two quantities are connected by a fixed rate and a fixed starting value. The solutions below explain each answer step by step using y = ax + b Class 9 methods.

Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.5 Solutions: Introduction to Linear Polynomials

Ex 2.5 Class 9 Ganita Manjari Chapter 2 Solutions

Ex 2.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 2.5 Solutions

Question 1.
A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹ 400. When she accessed 14 modules, her bill was ₹ 500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b.

Solution:

Given:
y = ax + b
When x = 10, y = 400:
400 = 10a + b ...(1)
When x = 14, y = 500:
500 = 14a + b ...(2)
Subtract equation (1) from equation (2):
500 - 400 = 14a - 10a
100 = 4a
a = 25
Substitute a = 25 in equation (1):
400 = 10(25) + b
400 = 250 + b
b = 150
Answer:
a = 25
b = 150
Therefore, the relation is:
y = 25x + 150

Question 2.
A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹ 800. When she used it for 15 hours, her bill was ₹ 1100. If the monthly billy depends on the hours of the use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.

Solution:

Given:
y = ax + b
When x = 10, y = 800:
800 = 10a + b ...(1)
When x = 15, y = 1100:
1100 = 15a + b ...(2)
Subtract equation (1) from equation (2):
1100 - 800 = 15a - 10a
300 = 5a
a = 60
Substitute a = 60 in equation (1):
800 = 10(60) + b
800 = 600 + b
b = 200
Answer:
a = 60
b = 200
Therefore, the relation is:
y = 60x + 200

Question 3.
Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C = a °F + b. Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit. (Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b, and thus, the linear relationship between °C and °F.)

Solution:

Given:
°C = a°F + b
When °C = 0 and °F = 32:
0 = 32a + b ...(1)
When °C = 100 and °F = 212:
100 = 212a + b ...(2)
Subtract equation (1) from equation (2):
100 = 180a
a = 100/180
a = 5/9
Substitute a = 5/9 in equation (1):
0 = 32(5/9) + b
b = -160/9
Therefore:
a = 5/9
b = -160/9
Hence, the linear relationship between Celsius and Fahrenheit is:
°C = (5/9)°F - 160/9
or,
°C = (5/9)(°F - 32)
Answer: a = 5/9 and b = -160/9.